$F/K$ finite extension, $p(x)$ irreducible with root in $F$, $\deg(p(x)) \mid [F:K]$?

Question

I am trying to give a reasoned answer to the following question:

Let $F/K$ be a finite field extension and let $p(x) \in K[x]$ be a irreducible polynomial (in $K[x]$, of course). If $p(x)$ has a root in $F$, is $[F:K]$ a multiple of $\deg(p(x))$?

I am pretty sure the answer to this question is $\textit{yes}$, but I am not sure how to make a formal argument for this. I have already proved that $[F:K] \neq \deg(p(x))$, but I don't know how to continue from here. Can someone help me?

Answer 1

Let $L$ be a rupture field of $p$ over $K$ and $a$ a root of $p$ in $F$. Then $[F:K]=[F:K(a)][K(a):K]$ as $p$ is supposed to be irreducible over $K$ and all rupture fields are isomorphic to $L$. In particular $K(a)$.

This indeed leads to a positive conclusion of the question as $\deg p = [K(a):K]$.

Answer 2

Let $a$ be a root of $p$ in $F$. Then $Irr(a,K)=p$ and $|K(a):K|=deg(p)$.

Hence $|F:K|=|F:K(a)|\cdot |K(a):K|=|F:K(a)|\cdot deg(p)$

Answer 3

Let $\alpha\in F$ be a root of $p(X)$. Then we have a tower of extensions $$ K\subset K[\alpha]\subset F $$ where $K[\alpha]$ is the smallest subfield of $F$ containing $K$ and $\alpha$. By the multiplicativity of the degree in towers $$ [F:K]=[F:K[\alpha]]\cdot[K[\alpha]:K] $$ and one concludes since $[K[\alpha]:K]=\deg(p(X))$.